Question

微积分的一个问题，涉及三角函数？

画圈的地方，是怎么消去的

Answer 1

x->0
cosαx = 1-(1/2)(αx)^2+o(x^2)
sinx.cosαx = x-(1/2)(αx)^2+o(x^2)
1+sinx.cosαx = 1+x-(1/2)(αx)^2+o(x^2)
ln(1+sinx.cosαx)
=ln[1+x-(1/2)(αx)^2+o(x^2)]
=[x-(1/2)(αx)^2+o(x^2)] -(1/2)[x-(1/2)(αx)^2+o(x^2)]^2 +o(x^2)
=[x-(1/2)(αx)^2+o(x^2)] -(1/2)[x^2+o(x^2)] +o(x^2)
=x +[ -(1/2)α^2 - 1/2 ]x^2 +o(x^2)
Similarly
ln(1+sinx.cosβx) =x +[ -(1/2)β^2 - 1/2 ]x^2 +o(x^2)
ln[ (1+ sinx.cosαx)/(1+sinx.cosβx)]
=ln(1+ sinx.cosαx)-ln(1+sinx.cosβx)
=[(1/2)β^2 -(1/2)α^2 ] x^2 +o(x^2)
//
lim(x->0) [ (1+ sinx.cosαx)/(1+sinx.cosβx)]^(cotx)^2
=lim(x->0) e^{ln[ (1+ sinx.cosαx)/(1+sinx.cosβx)] / (tanx)^2 }
=lim(x->0) e^{ln[ (1+ sinx.cosαx)/(1+sinx.cosβx)] / x^2 }
=lim(x->0) e^{ [(1/2)β^2 -(1/2)α^2 ] x^2 / x^2 }
=e^ [(1/2)β^2 -(1/2)α^2 ]